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T W E L V E
Design via State Space
SOLUTION TO CASE STUDY CHALLENGE
Antenna Control: Design of Controller and Observer
a. We first draw the signal-flow diagram of the plant using the physical variables of the system
as state variables.
Writing the state equations for the physical variables shown in the signal-flow diagram, we
obtain
0 1 0 0
0 -1.32 0.8 0 u ; y [0.2 0 0]
0 0 -100 2000
z z z
The characteristic polynomial for this system is s3 + 101.32s2 + 132s + 0. Hence, the A and B
matrices of the phase-variable form are
0 1 0 0
0 0 1 0
0 -132 -101.32 1
Ax Bx
Writing the controllability matrices and their determinants for both systems yields
12-2 Chapter 12: Design via State Space
Controllability Matrix of z Controllability Matrix of x
0 0 1600 0 0 1
0 1600 -162112 0 1 -101.32
2000 -200000 20000000 1 -101.32 10133.7424
-5.12 E 09 -1
CMz CMx
Det(CMz) Det(CMx)
where the system is controllable. Using Eq. (12.39), we find the transformation matrix and its inverse
to be
Transformation Matrix z Px
1600 0 0 0.000625 0 0
0 1600 0 0 0.000625 0
0 2640 2000 0 -0.000825 0.0005
P PINV
The characteristic polynomial of the phase-variable system with state feedback is
s3 + (k3 + 101.32)s
2 + (k2 + 132)s + (k1 + 0)
For 15% overshoot, Ts = 2 seconds, and a third pole 10 times further from the imaginary axis than
the dominant poles, the characteristic polynomial is
(s + 20)(s2 + 4s + 14.969) = s3 + 24s2 + 94.969s + 299.38
Equating coefficients, the controller for the phase-variable system is
Controller for x
299.38 -37.031 -77.32
Kx
Using Eq. (12.42), the controller for the original system is
Controller for z
0.1871125 0.04064463 -0.03866
Kz
b. Using Kz, gain from m = - 0.1871125 (including gear train, pot, and operational amplifier); gain
from tachometer = - 0.04064463; and gain from power amplifier output = 0.03866.
Solution to Case Study Challenge 12-3
c. Using the original system from part (a) and its characteristic polynomial, we find the observer
canonical form which has the following A and C matrices:
-101.32 1 0
-132 0 1
0 0 0
1 0 0
Ax
Cx
Writing the observability matrices and their determinants for both systems yields
Observability Matrix of z Observability Matrix of x
0.2 0 0 1 0 0
0 0.2 0 -101.32 1 0
0 -0.264 0.16 10133.7424 -101.32 1
0.0064 1
OMz OMx
Det(OMz) Det(OMx)
where the system is observable. Using Eq. (12.89), we find the transformation matrix and its inverse to be
Transformation Matrix z Px
5 0 0 0.20 0.00 0.00
-506.6 5 0 20.26 0.20 0.00
62500 -625 6.25 26.40 20.00 0.16
P PINV
The characteristic polynomial of the dual phase-variable system with state feedback is
s3 + (l1 + 101.32)s
2 + (l2 + 132)s + (l3 + 0)
12-4 Chapter 12: Design via State Space
For 10% overshoot, n = 10 14.969 = 38.69 rad/s, and a third pole 10 times further from the
imaginary axis than the dominant observer poles, the characteristic polynomial is
(s + 228.72)(s2 + 45.743s + 1496.916) = s3 + 274.46s2 + 11959s + 3.4237x105
Equating coefficients, the observer for the observer canonical system is
Observer for x
173.14
11827
342370
Lx
Using Eq. (12.92), the observer for the original system is
Observer for z
865.7
-28577.724
5569187.5
Lz
d.
e.
Program:
'Controller'
Solution to Case Study Challenge 12-5
A=[0 1 0;0 -1.32 0.8;0 0 -100];
B=[0;0;2000];
C=[0.2 0 0];
D=0;
pos=input('Type desired %OS ');
Ts=input('Type desired settling time ');
z=(-log(pos/100))/(sqrt(pi^2+log(pos/100)^2));
wn=4/(z*Ts); %Calculate required natural
%frequency.
[num,den]=ord2(wn,z); %Produce a second-order system that
%meets the transient response
%requirements.
r=roots(den); %Use denominator to specify dominant
%poles.
poles=[r(1) r(2) 10*real(r(1))];
%Specify pole placement for all
%poles.
K=acker(A,B,poles)
'Observer'
pos=input('Type desired %OS ');
z=(-log(pos/100))/(sqrt(pi^2+log(pos/100)^2));
wn=10*wn %Calculate required natural
%frequency.
[num,den]=ord2(wn,z); %Produce a second-order system that
%meets the transient response
%requirements.
r=roots(den); %Use denominator to specify dominant
%poles.
poles=[r(1) r(2) 10*real(r(1))];%Specify pole placement for all
%poles.
l=acker(A',C',poles)'
Computer response:
ans =
Controller
Type desired %OS 15
Type desired settling time 2
K =
0.1871 0.0406 -0.0387
ans =
Observer
Type desired %OS 10
wn =
38.6899
l =
12-6 Chapter 12: Design via State Space
1.0e+006 *
0.0009
-0.0286
5.5691
ANSWERS TO REVIEW QUESTIONS
1. Both dominant and non-dominant poles can be specified with state-space design techniques.
2. Feedback all state variables to the plant's input through a variable gain for each. Decide upon a closed-
loop characteristic equation that has a pole configuration to yield a desired response. Write the
characteristic equation of the actual system. Match coefficients and solve for the values of the variable
gains.
3. Phase-variable form
4. The control signal developed by the controller must be able to affect every state variable.
5. If the signal-flow diagram is in the parallel form, which leads to a diagonal system matrix, controllability
can be determined by inspection by seeing that all state variables are fed by the control signal.
6. The system is controllable if the determinant of the controllability matrix is non-zero.
7. An observer is a system that estimates the state variables using information from the output of the actual
plant.
8. If the plant's state-variables are not accessible, or too expensive to monitor
9. An observer is a copy of the plant. The difference between the plant's output and the observer's output is
fed back to each of the derivatives of the observer's state variables through separate variable gains.
10. Dual phase-variable
11. The characteristic equation of the observer is derived and compared to a desired characteristic equation
whose roots are poles that represent the desired transient response. The variable gains of each feedback path
are evaluated to make the coefficients of the observer's characteristic equation equal the coefficients of the
desired characteristic equation.
12. Typically, the transient response of the observer is designed to be much faster than that of the controller.
Since the observer emulates the plant, we want the observer to estimate the plant's states rapidly.
13. Det[A-BK], where A is the system matrix, B is the input coupling matrix, and K is the controller.
14. Det[A-LC], where A is the system matrix, C is the output coupling matrix, and L is the observer.
15. The output signal of the system must be controlled by every state variable.
16. If the signal-flow diagram is in the parallel form, which leads to a diagonal system matrix, observability
can be determined by inspection by seeing that all state variables feed the output.
17. The system is observable if the determinant of the observability matrix is non-zero.
Solutions to Design Problems 12-7
SOLUTIONS TO PROBLEMS
1.
i.
2 2
( 3) 1
( ) *( 3)
( 4) 8 16
s
G s s
s s s
a.
b.
c.
12
0 1 0
; 3 1
( 16) ( 8) 1
r y
k k
x x x
d.
12-8 Chapter 12: Design via State Space
2 2 1
3
( )
( 8) ( 16)
s
T s
s k s k
1 1 1 2 2 3 3
2 1 1 2 2 2 3 3
3 1 1 2 2 3 3
1 2 3
1 2 2 3
1 2
( 20 71.25 ) 71.25 71.25 71.25
27.5 ( 10 27.5 ) 27.5 27.5
6.25 6.25 6.25 6.25
( 20 71.25 ) 71.25 71.25
27.5 ( 10 27.5 ) 27.5
6.25 6.25 6
k x k x k x r
k x x k x k x r
k x k x k x r
k k k
k x k k
k k
x
x
x
A
3
2
3 2
1 2 3 1 2 3 3
71.25
; 27.5 ; 1 1 1
.25 6.25
200( 7 25)
( )
4 (120 285 110 25 ) (800 2850 2200 750 ) 5000
k
s s
T s
s k k k s k k k s k
B C
e.
Part d. yields same result as i(d).
ii.
2
1
( ) *
( 5)( 7) 12 35
s
G s s
s s s s
a.
b.
Solutions to Design Problems 12-9
c.
1 2
0 1 0
; 0 1
( 35) ( 12) 1
r y
k k
x x x
d.
2 2 1
( )
( 12) ( 35)
s
T s
s k s k
e.
1
1 2
0 1 0
( ) (s ) ; ; ; 0 1
( 35) ( 12) 1
T s
k k
C I A B A B C
which yields the same result as ii(d).
iii.
2
3 2
20 ( 7) 1
( ) = = * (20s +140s)
( 3)( 7)( 9) 19 111 189
s s
G s
s s s s s s
a.
b.
12-10 Chapter 12: Design via State Space
c.
1 2 3
0 1 0 0
0 0 1 0 ; 0 140 20
( 189) ( 111) ( 19) 1
r y
k k k
x x x
d.
3 23 2 1
20 ( 7)
( ) =
( 19) ( 111) ( 189)
s s
T s
s k s k s k
e.
1
1 2 3
0 1 0 0
( ) (s ) ; 0 0 1 ; 0 ; 0 140 20
( 189) ( 111) ( 19) 1
T s
k k k
C I A B A B C
which yields the same result as iii(d).
iv.
2
3 2
30( 2)( 3) 1
( ) = = * (30s 150 180)
( 4)( 5)( 6) 15 74 120
s s
G s s
s s s s s s
Solutions to Design Problems 12-11
a.
b.
c.
1 2 3
0 1 0 0
0 0 1 0 ; 180 150 30
( 120) ( 74) ( 15) 1
r y
k k k
x x x
d.
12-12 Chapter 12: Design via State Space
2
3 2
3 2 1
30 150 180
( ) =
( 15) ( 74) ( 120)
s s
T s
s k s k s k
e.
1
1 2 3
0 1 0 0
( ) (s ) ; 0 0 1 ; 0 ; 180 150 30
( 120) ( 74) ( 15) 1
T s
k k k
C I A B A B C
which yields the same result as iv(d).
v.
2
2
2 2 4 3 2
s 8 15 1
( ) = = * (s 8 15)
(s 4 10)(s 3 12) 7 34 78 120
s
G s s
s s s s s s
a.
b.
c.
Solutions to Design Problems 12-13
1 2 3 4
0 1 0 0 0
0 0 1 0 0
; 15 8 1 0
0 0 0 1 0
( 120) ( 78) ( 34) ( 7) 1
r y
k k k k
x x x
d.
2
4 3 2
4 3 2 1
8 15
( )
( 7) ( 34) ( 78) ( 120)
s s
T s
s k s k s k s k
e.
1
1 2 3 4
0 1 0 0 0
0 0 1 0 0
( ) (s ) ; ; ; 15 8 1 0
0 0 0 1 0
( 120) ( 78) ( 34) ( 7) 1
T s
k k k k
C I A B A B C
which yields the same result as v(d).
2.
i
a. The output is
Since,
2
1 1 1 1 2 2 1
2 3 2 1 1 2 3
(30 270 420) 30 270 420 30 270 420
30( 5 ) 270 420 420 120 30
y s s x x x x x x x
x x x x x x x
12-14 Chapter 12: Design via State Space
b.
1
1 2 3
0 1 0 0
( ) (s ) ; 0 5 1 ; 0 ; 420 120 30
( 3) 1
T s
k k k
C I A B A B C
3 23 3 2 1
30( 2)( 7)
( )
( 8) (5 15)
s s
T s
s k s k k s k
ii
a.
b.
1
1 2 3
0 1 0 0
T(s) (s ) ; 0 0 1 ; 0 ; -3 1 1
k k (5k 2) 5
C I A B A B C
2
3 2
3 3 2 3 1
5(s 3s 7)
s (5k 4)s (10k k 14)s (50k k 20)
3.
i
a.
Solutions to Design Problems 12-15
250( 7 25) 6.25 27.5 71.25
( )
( 10)( 20) 10 20
s s
G s
s s s s s s
b. Writing the state equations:
1 1
2 2
3
20 71.25
10 27.5
6.25
x u
x u
u
x
x
x
But, u = -k1x1 - k2x2 - k3x3 + r . Substituting into the state equations,
1 1 1 2 2 3 3
2 1 1 2 2 2 3 3
3 1 1 2 2 3 3
( 20 71.25 ) 71.25 71.25 71.25
27.5 ( 10 27.5 ) 27.5 27.5
6.25 6.25 6.25 6.25
k x k x k x r
k x x k x k x r
k x k x k x r
x
x
x
Therefore, T(s) = C(sI - A)-1B, where
1 2 3
1 2 2 3
1 2 3
( 20 71.25 ) 71.25 71.25 71.25
27.5 ( 10 27.5 ) 27.5 ; 27.5 ; 1 1 1
6.25 6.25 6.25 6.25
k k k
k x k k
k k k
A B C
Hence,
2
3 2
1 2 3 1 2 3 3
200( 7 25)
( )
4 (120 285 110 25 ) (800 2850 2200 750 ) 5000
s s
T s
s k k k s k k k s k
12-16 Chapter 12: Design via State Space
ii
a.
50( 3)( 4) 50 300 300
( )
( 5)( 6)( 7) 5 6 7
s s
G s
s s s s s s
b. Writing the state equations:
1 1
2 2
3 3
5 50
6 300
7 300
x x u
x x u
x x u
But,
1 1 2 2 3 3u k x k x k x r
Substituting into the state equations, collecting terms, and converting to vector-matrix form yields
1 2 3
1 2 3
1 2 3
(5 ) 50 50 50
300 (300 6) 300 300
(300 7) 300 300 300
1 1 1
k k k
k k k r
k k k
y
x x
x
Therefore, T(s) = C(sI - A)-1B, or
2
1
3 2
3 2 1 3 2 1 3 1 2 1
50 1750 (6900 88200 )
( )
(300 300 11) 2(1475 750 3 7350 7350 15)
3
s s k
T s
s k k k s k k k k k k k s
3 100 (23 294 )k k
Solutions to Design Problems 12-17
4.
The plant is given by
3 2
50 50
( )
( 1)( 3)( 10) 14 43 30
G s
s s s s s s
The characteristic polynomial for the plant with phase-variable state feedback is
3 23 2 1( 14) ( 43) ( 30) 0s k s k s k
The desired characteristic equation is
2 3 2( 80)( 16 183.137) 96 1463 14650s s s s s s
based upon 10% overshoot, 0.5sT second, and a third pole ten times further from the
imaginary axis than the dominant poles. Comparing both characteristic equations
1 2 314620, 1420, 82k k k
5.
a. The system in controller canonical form is:
A B C
n 1 n 2 1 0
1 2 3 4
a a a a 1
1 0 0 0 0
= ; = ; = c c c c
0
0 0 1 0 0
The characteristic equation of the plant is:
sn + an-1s
n-1 + . . . + a1s + a0 = 0
Forming the closed-loop system by feeding back each state variable and the input to u forming
u = -Kx + r
where
K = [k1 k2 . . . kn]
and substituting u into the state equation, we obtain
u ( ) r x Ax B A BK x B= + =
Forming A - BK:
12-18 Chapter 12: Design via State Space
1 2 n 1 n(a k ) (a k ) (a k ) (a k )n 1 n 2 1 0
1 0 0 0
0 0 1 0
A BK
The characteristic equation is:
sn + (an-1 + k1)s
n-1 + (an-2 + k2). . . + (a1 + kn-1)s + (a0 + kn) = 0
Assuming a desired characteristic equation,
sn + dn-1s
n-1 + dn-2s
n-2+ . . . + d2s
2 + d1s + d0 = 0
Equating coefficients,
di = ai + kn-i ; i = 0, 1, 2, ... n-1
from which
kn-i = di – ai (1)
b. The desired characteristic equationis
3 215.9 136.08 413.1 0s s s
the characteristic equation of the plant is
3 25 4 0 0s s s
Using Eq. (1) above, k3-i = di – ai. Therefore, k3 = d0 – a0 = 413.1 – 0 = 413.1; k2 = d1 – a1 = 136.08 –
4 = 132.08; k1 = d2 – a2 = 15.9 – 5 = 10.9. Hence,
10.9 132.08 413.1K
6.
Using Eqs. (4.39) and (4.34), we find ζ= 0.5192 and
2 1/ 2
7.352
(1 )
n
PT
, respectively.
Factoring the denominator of Eq. (4.22), the required poles are – 3.676 ± j6.2832. We place the third
pole at -2 to cancel the open loop zero. Multiplying the three closed-loop pole terms yields the desired
characteristic equation:
s
3
+ 9.6343 s
2
+ 69.3186 s + 108.1 = 0.
Since
2
3 2
100 2700 5000
( )
9 23 15
s s
G s
s s s
, the controller canonical form is:
A =
9 23 15
1 0 0
0 1 0
; B =
1
0
0
; C = 100 2700 5000 ;
Solutions to Design Problems 12-19
The first row of A contains the coefficients of the characteristic equation. Thus comparing the first
row of A to the desired characteristic equation and using the results of Problem 5, we have:
k1 = - (9.6343 - 9) = 0.6343; k2 = - (69.3186 - 23) = 46.3186; and k3 = - (108.1 - 15) = 93.1.
7.
The plant is given by
3 2
20( 2) 20 40
( )
( 5)( 7) 12 35 0
s s
G s
s s s s s s
The characteristic polynomial for the plant with phase-variable state feedback is
3 23 2 3( 12) ( 35) ( 0)s k s k s k
The desired characteristic equation is
2 3 2( 20)( 4 11.45) 24 91.45 229s s s s s s
based upon 10% overshoot, Ts =2 seconds, and a third pole ten times further from the imaginary axis
than the dominant poles. Comparing the two characteristic equations,
1 2229, 56.45,k k and 3k 12 .
8.
Drawing the signal-flow diagram,
Writing the state equations yields the following A matrix:
1 2 3
0 1 0
0 4 1
[2 ]
A
k k k
from which,
3 23 3 2 1| S | ( 6) (4 8)s k s k k s kI A
12-20 Chapter 12: Design via State Space
The desired characteristic equation is (s + 80)(s2 + 16s + 183.137) = s3 + 96s2 + 1463.1s + 14651
based upon 10% overshoot, Ts = 0.5 second, and a third pole ten times further from the imaginary axis
than the dominant poles. Comparing the two characteristic equations, , k1 = 14651, k2 = 1095.1, and
k3 = 90.
9.
Expand G(s) by partial fractions and obtain
20 1.67 5 6.67
( )
( 4)( 6) 4 6
G s
s s s s s s
Drawing the signal-flow diagram with state feedback
Writing the state equations yields the following system matrix:
1 2 3
1 2 3
1 2 3
1.67 1.67 1.67
5 (5 4) 5
6.67 6.67 (6.67 6)
k k k
k k k
k k k
A
Evaluating the characteristic polynomial yields,
23 2 1 3 2 1 1( 6.67 5 1.67 10) ( 26.68 30 16.7 24) 40.08s k k k s k k k s kI A
From Problem 7, the desired characteristic polynomial is
3 224 91.45 229s s s .
Equating coefficients and solving simultaneously yields
1 25.71, 4.58,k k and 3k 4.10 .
10.
Solutions to Design Problems 12-21
a.
Writing the state equation and the controllability matrix for the system yields
1
2
7 1
1 2
b
u
b
x x
1 1 2
2 1 2
7
2
b b b
b b b
MC B AB
The controllability matrix has a zero determinant if
2 21 1 2 25 0b b b b
b.
If 1 1b the condition becomes
2
2 25 1 0b b so there are two possible solutions 2 0.2087b
and 2 4.7913b .
11.
The controllability matrix is given by Eq. (12.26) for each of the following solutions:
a.
-2 0 1 0 0 1 -5
0 -2 0 ; 1 ; 1 -2 4 ;det 0 ; system is uncontrollable
0 0 -3 1 1 -3 9
M MA B C C
b.
-2 1 0 0 0 1 -4
0 -2 0 ; 1 ; 1 -2 4 ;det 1; system is controllable
0 0 -3 1 1 -3 9
M MA B C C
c.
-4 1 0 0 0 2 -7
0 0 1 ; 2 ; 2 1 -3 ;det 7 ; system is controllable
0 0 -3 1 1 -3 9
M MA B C C
d.
-4 1 0 1 1 -4 17
0 0 1 ; 0 ; 0 1 -8 ;det 5 ; system is controllable
-5 0 -3 1 1 -8 44
M MA B C C
12-22 Chapter 12: Design via State Space
e.
0 1 1 1 -2
; ; ;det 0 ; system is uncontrollable
-6 -5 -2 -2 4
M MA B C C
f.
-4 0 0 1 1 -4 16
0 -5 0 ; 0 ; 0 0 0 ;det 0 ; system is uncontrollable
0 0 -6 1 1 -6 36
M MA B C C
This system can also be determined uncontrollable by inspection.
12.
Program:
'(d)'
A=[-4 1 0;0 0 1;-5 0 -3]
B=[1;0;1]
Cm=ctrb(A,B)
Rank=rank(Cm)
pause
'(f)'
A=[-4 0 0;0 -5 0;0 0 -6]
B=[1;0;1]
Cm=ctrb(A,B)
Rank=rank(Cm)
Computer response:
ans =
(d)
A =
-4 1 0
0 0 1
-5 0 -3
B =
1
0
1
Cm =
1 -4 17
0 1 -8
1 -8 44
Rank =
3
ans =
(f)
A =
Solutions to Design Problems 12-23
-4 0 0
0 -5 0
0 0 -6
B =
1
0
1
Cm =
1 -4 16
0 0 0
1 -6 36
Rank =
2
13.
From Eq. (12.46) we write the controller canonical form:
ccA
8 17 10
= 1 0 0
0 1 0
,
ccB
1
= 0
0
.
The controllability matrices are found using Eq. (12.35). For the original system of Eq. (12.44),
MzC
0 0 1
= 0 1 3
0 1 1
. For the controller canonical form,
MccC
1 8 47
= 0 1 8
0 0 1
. The transformation matrix
is, P = CMzCMcc
-1 =
0 0 1
0 1 5
1 7 10
. Comparing the first row of Acc with the desired characteristic
equation, Eq. (12.50), Kcc = [ -2 -4 10]. Transforming back to the original system, Kz = KccP
-1 =
[-20 10 -2].
14.
Drawing the signal-flow diagram for the plant in cascade form yields
Writing the A and B matrices for the z system,
Az Bz
-3 1 0 0
0 -8 1 0
0 0 -10 1
12-24 Chapter 12: Design via State Space
Writing the A and B matrices for the x (phase-variable) system,
Ax Bx
0 1 0 0
0 0 1 0 Phase-Variable Form
-240 -134 -21 1
From the phase variable from, the characteristic polynomial is s3 + 21s2 + 134s + 240.
Finding the controllability matrices and their determinants for the z and x systems shows that there is
controllability,
CMz Controllability Matrix of z CMx Controllability Matrix of x
0 0 1 0 0 1
0 1 -18 0 1 -21
1 -10 100 1 -21 307
Det(CMz) -1 Det(CMx) -1
Using Eq. (12.39), the transformation matrix P and its inverse are found to be
P Transformation Matrix z=Px PINV
1 0 0 1.00 0.00 0.00
3 1 0 -3.00 1.00 0.00
24 11 1 9.00 -11.00 1.00
Using the given transient requirements, and placing the third closed-loop pole over the zero at -6
yields the following desired closed-loop characteristic polynomial:
(s2 + 8s + 45.78)(s + 6) = s3 + 14s2 + 93.78s + 274.68
Using the phase-variable system with state feedback the characteristic polynomial is
s3 + (k3 + 21)s
2 + (k2 + 134)s + (k1 + 240)
Equating the two characteristic polynomials yields the state feedback vector for the x system as
Kx Controller for x
34.68 -40.22 -7
Using Eq. (12.42),
Kz Controller for z
92.34 36.78 -7
15.
Program:
A=[-3 1 0;0 -8 1;0 0 -10]; %Generate system matrix A
B=[0;0;1]; %Generateinput coupling matrix B
C=[3 1 0]; %Generate output coupling matrix C
D=0; %Generate matrix D
Solutions to Design Problems 12-25
Po=10; %Input desired percent overshoot
Ts=1; %Input desired settling time
z=(-log(Po/100))/(sqrt(pi^2+log(Po/100)^2));
%Calculate required damping ratio
wn=4/(z*Ts); %Calculate required natural
%frequency
[num,den]=ord2(wn,z); %Produce a second-order system that
%meets transient requirements
r=roots(den); %Use denominator to specify
%dominant poles
poles=[r(1) r(2) -6]; %Specify pole placement for all
%poles.
%A few tries at the the third-pole
%value shows T(s) with a closed-
%loop zero at -7.
%Thus, choose the third pole to
%cancel this zero.
K=acker(A,B,poles) %Calculate controller gains in z-
%system
Anew=A-B*K; %Form compensated A matrix
Bnew=B; %Form compensated B matrix
Cnew=C; %Form compensated C matrix
Dnew=D; %Form compensated D matrix
[numt,dent]=ss2tf(Anew,Bnew,Cnew,Dnew);
%Form T(s)
'T(s)' %Display label
T=tf(numt,dent) %Display T(s)
poles=pole(T) %Display poles of T(s)
Computer response:
K =
92.3531 36.7844 -7.0000
ans =
T(s)
Transfer function:
-3.553e-015 s^2 + s + 6
------------------------------
s^3 + 14 s^2 + 93.78 s + 274.7
poles =
-4.0000 + 5.4575i
-4.0000 - 5.4575i
-6.0000
16.
Expanding by partial fractions,
( 6) 0.085714 0.2 0.28571
( )
( 3)( 8)( 10) ( 3) ( 8) ( 10)
s
G s
s s s s s s
Writing the A and B matrices for the z system with ki's set to zero,
Az Bz
-3 0 0 0.085714
0 -8 0 0.2
12-26 Chapter 12: Design via State Space
0 0 -10 -0.28571
Writing the A and B matrices for the x (phase-variable) system,
Ax Bx
0 1 0 0
0 0 1 0 Phase-Variable Form
-240 -134 -21 1
From the phase variable from, the characteristic polynomial is s3 + 21s2 + 134s + 240.
Finding the controllability matrices and their determinants for the z and x systems shows that there is
controllability,
CMz Controllability Matrix of z CMx Controllability Matrix of x
0.085714 -0.257142 0.771426 0 0 1
0.2 -1.6 12.8 0 1 -21
-0.28571 2.8571 -28.571 1 -21 307
Det(CMz) 0.342850857 Det(CMx) -1
Using Eq. (12.39), the transformation matrix P and its inverse are found to be
P Transformation Matrix z=Px PINV
6.85712 1.542852 0.085714 0.33 -0.50 -0.25
6 2.6 0.2 -1.00 4.00 2.50
-6.85704 -3.14281 -0.28571 3.00 -32.00 -25.00
Using the given transient requirements, and placing the third closed-loop pole over the zero at -6
yields the following desired closed-loop characteristic polynomial:
(s2 + 8s + 45.78)(s + 6) = s3 + 14s2 + 93.78s + 274.68
Using the phase-variable system with state feedback the characteristic polynomial is
s3 + (k3 + 21)s
2 + (k2 + 134)s + (k1 + 240)
Equating the two characteristic polynomials yields the state feedback vector for the x system as
Kx Controller for x
34.7062 -40.2156 -7
Using Eq. (12.42),
Kz Controller for z
30.78443595 45.7845 65.78543678
17.
Solutions to Design Problems 12-27
Draw signal-flow diagram showing state variables, z, at the output of each subsystem and the state
variables, w, at the output of the integrators.
Recognizing that 2 2 2 3 2 36 8 2z w w w w w , we can write the state equations for w as
10 2 1 0
0 8 1 0
0 0 3 1
1 0 0
u
y
w w
w
Writing the relationship between z and w yileds
1
1 0 0
0 2 1
0 0 1
z w P w
Thus
1 0 0
0 0.5 0.5
0 0 1
P
Converting the state equations in w to state equations in z, we use Eqs. (5.87) and obtain the A matrix
and B vector as
Az Bz
-10 1 0 0
0 -8 3 1
0 0 -3 1
Writing the A and B matrices for the x (phase-variable) system,
Ax Bx
0 1 0 0
0 0 1 0 Phase-Variable Form
-240 -134 -21 1
From the phase variable from, the characteristic polynomial is s3 + 21s2 + 134s + 240
Finding the controllability matrices and their determinants for the z and x systems shows that there is
controllability,
12-28 Chapter 12: Design via State Space
CMz Controllability Matrix of z CMx Controllability Matrix of x
0 1 -15 0 0 1
1 -5 31 0 1 -21
1 -3 9 1 -21 307
Det(CMz) -8 Det(CMx) -1
Using Eq. (12.39), the transformation matrix P and its inverse are found to be
P Transformation Matrix z=Px PINV
6 1 0 -0.25 -0.13 0.13
60 16 1 2.50 0.75 -0.75
80 18 1 -25.00 -3.50 4.50
Using the given transient requirements, and placing the third closed-loop pole over the zero at -6
yields the following desired closed-loop characteristic polynomial:
(s2 + 8s + 45.78)(s + 6) = s3 + 14s2 + 93.78s + 274.68
Using the phase-variable system with state feedback the characteristic polynomial is
s3 + (k3 + 21)s
2 + (k2 + 134)s + (k1 + 240)
Equating the two characteristic polynomials yields the state feedback vector for the x system as
Kx Controller for x
34.68 -40.22 -7
Using Eq. (12.42),
Kz Controller for z
65.78 -10 3
18.
Using Eqs. (4.39) and (4.34), we find = 0.456 and n = 17.6498. Factoring the
denominator of Eq. (4.22), the required poles are – 8.0483 ± j15.708. We place the third
pole 10 times further at – 80.483. Multiplying the three closed-loop pole terms yields the
desired characteristic equation: 3 296.6 1607 25072 0s s s . Representing the plant
in parallel form:
0 0 0 2.5
0 4 0 ; 4.1667 ; 1 1 1 ;
0 0 10 1.667
par par parA B C
Using Eq. (12.26),
2.5 0 0
4.1667 16.667 66.667 ,
1.667 16.667 166.7
M parC which is controllable since the
determinant is 4168.1.
Solutions to Design Problems 12-29
Since,
3 2
100
( )
14 40
G s
s s s
the controller canonical form is:
14 40 0 1
1 0 0 ; 0 ; 0 0 100
0 1 0 0
cc ccA B C . Using Eq. (12.26),
1 14 156
0 1 14
0 0 1
MccC , which is controllable since the determinant is 1.
2.5 0 0
4.1667 16.667 66.667 ,
1.667 16.667 166.7
M parC
The first row of ACC contains the coefficients of the characteristic equation. Comparing the
first row of ACC to the desired characteristic equation and using the results of Problem 5, we
have:
(14 + k1) = 96.6; (40 + k2) = 1607; and (0 + k3) = 25072. Hence, KCC = [25072 1567
82.6].
The transformation matrix is:
2.500 35.00 100.0
4.1667 41.6668 0.0030
1.6670 6.6710 0.0420
1
M par MccP C C
Transforming back to the original system: 10.0 3942 2490.81par CCK K P
19.
a.
>> A=[-0.014 0 -1.4 0 0; 0.023 -0.023 -0.023 0 0; 0.134 0.67 -0.67 0.38 0.003264; 0 0 0.06 -0.06 0;
0 0 0.0017 0 -0.001];
>> B = [1;0;0;0;0];
>> rank(ctrb(A,B))
12-30 Chapter 12: Design via State Space
ans =
5
The controllability matrix has full rank so the system is controllable.
b.
>> [P,Ad]=jordan(A)% P is the similarity transformation matrix used for diagonalization, Ad the
diagonalized A matrix
P =
-0.0000 + 0.0000i 1.3181 + 0.1838i 0.0446 -1.6808 1.3181 - 0.1838i
-0.0000 + 0.0000i -0.0336 - 0.0948i -0.0286 0.0958 -0.0336 + 0.0948i
0.0000 - 0.0000i 0.1916 - 0.1798i 0.0014 -0.3845 0.1916 + 0.1798i
0.0000 - 0.0000i -0.0605 - 0.0176i 0.0370 0.0841 -0.0605 + 0.0176i
0.0012 - 0.0000i -0.0016 - 0.0002i -0.0000 0.0020 -0.0016 + 0.0002i
Ad =
-0.0010 0 0 0 0
0 -0.1875 + 0.2151i 0 0 0
0 0 -0.0577 0 0
0 0 0 -0.3343 0
0 0 0 0 -0.1875 - 0.2151i
Solutions to Design Problems 12-31
>> Bd=inv(P)*B
Bd =
1.0000 + 0.0000i
1.0000
1.0000 + 0.0000i
1.0000 + 0.0000i
1.0000 + 0.0000i
20.
3 2
1 1
( )
( 2)( 5) 7 10
G s
s s s s s s
The corresponding A and C matrices in observer canonical form are
7 1 0
10 0 1
0 0 0
A
1 0 0C
The characteristic polynomial is 3 27 10 0s s s . We check observability by
calculating the observability matrix and its determinant
1 0 0
7 0 1
39 7 1
M
2
C
O CA
CA
| | 1MO
Using the given transient requirements, and placing the third closed-loop pole 10 times
farther from the imaginary axis than the dominant pole yield the following desired
characteristic polynomial
2 3 2( 300)( 60 2500) 360 20500 750000s s s s s s
Equating this polynomial to the one in Equation (12.67) yields the observer gains
12-32 Chapter 12: Design via State Space
1 2 3353, 20490, 750000l l l
21.
Using Eqs. (4.39) and (4.34) to find 0.5912 and n = 19.4753 respectively. Factoring the
denominator of Eq. (4.22), the required poles are -11.513 ± j15.708. We place the third pole 20 times
further at –230.26. Multiplying the three closed-loop pole terms yields the desired characteristic
equation: s3 + 253.28s2 + 5681.19s + 87334.19 = 0.
Representing the plant in observer canonical form:
25 1 0
171 0 1
315 0 0
ocA ;
0
0
10
ocB ;
Coc = [1 0 0]. The first column of Aoc contains the coefficients of the characteristic equation.
Comparing the first column of Aoc to the desired characteristic equation and using Eq. (12.67), l1 =
253.28 - 25 = 228.28; l2 = 5681.19 - 171 = 5510.19; and l3 = 87334.19 - 315 = 87019.19. Hence,
Loc = [228.28 5510.19 87019.19]
T.
22.
The A, L, and C matrices for the phase-variable system are:
0 1 0
0 0 1
0 21 10
A
1 0 0C
1
2
3
l
l
l
L
Hence,
1
2
3
1 0
( ) 1
21 10
l
l
l
A LC
or
3 21 1 2 1 2 3( ) (10 ) (21 10 ) (21 10 )l l l l l lA LC
From Problem 19, the desired characteristic polynomial is 3 + 3602 + 23625+ 1687500.
Equating coefficients yields:
1 1 2 1 2 310 360; (21 10 ) 23625; (21 10 ) 1687500l l l l l l
Solutions to Design Problems 12-33
Solving successively,
1 2 3350; 20104; 1479110l l l
23.
The A, L, and C matrices for the phase-variable system are:
1
2
0 1
; 2 1 ; =
45 14
l
l
A C L
Hence,
1 1
2 2
2 1
( )
2 45 14
l l
l l
A LC
or
2 1 2 2 1(2 14) (2 17 45)l l l l
From the problem statement, the desired characteristic polynomial is 2 + 144+ 14400.
Equating coefficients yields,
1 2 2 1(2 14) 144; (2 17 45) 14400l l l l
Solving simultaneously,
1 2671.2; 1472.4l l
24.
The A matrix for each part is given in the solution to Problem 11. Each observability matrix is
calculated from Eq. (12.79).
a.
2 0 1 5 5 5
0 2 0 ; (5,5,5) ; 10 10 10 ; | | 0 ; unobservable
0 0 3 20 20 20
M MA C O O
b.
2 1 0 5 0 5
0 2 0 ; (5,0,5) ; 10 5 15 ; | | 125 ; observable
0 0 3 20 20 45
M MA C O O
c.
4 1 0 1 0 0
0 0 1 ; (1,0,0) ; 4 1 0 ; | | 1; observable
0 0 3 16 4 1
M MA C O O
d.
4 1 0 1 0 0
0 0 1 ; (1,0,0) ; 4 1 0 ; | | 1; observable
5 0 3 16 4 1
M MA C O O
e.
0 1 1 0
; C (1,0) ; ; | | 1; observable
6 5 0 1
M MA O O
f.
12-34 Chapter 12: Design via State Space
4 0 0 1 1 1
0 5 0 ; (1,1,1) ; 4 5 6 ; | | 2 ; observable
0 0 6 16 25 36
M MA C O O
25.
Program:
'(a)'
A=[-2 0 1;0 -2 0;0 0 -3] %Form compensated A matrix
C=[5 5 5] %Form compensated C matrix
Om=obsv(A,C) %Form observability matrix
Rank=rank(Om) %Find rank of observability
%matrix
'(f)'
A=[-4 0 0;0 -5 0;0 0 -6] %Form compensated A matrix
C=[1 1 1] %Form compensated C matrix
Om=obsv(A,C) %Form observability matrix
Rank=rank(Om) %Find rank of observability
Computer response:
ans =
(a)
A =
-2 0 1
0 -2 0
0 0 -3
C =
5 5 5
Om =
5 5 5
-10 -10 -10
20 20 20
Rank =
1
ans =
(f)
A =
-4 0 0
0 -5 0
0 0 -6
C =
1 1 1
Om =
1 1 1
-4 -5 -6
16 25 36
Rank =
Solutions to Design Problems 12-35
3
26.
a.
Representing the system in steady state yields
0 1 0
2 4 1
ux x
1 2y c c x
The observability matrix is
1 2
2 1 22 4
c c
c c c
M
C
O
CA
It’s determinant is
2 21 1 2 2| | 4 2c c c cMO
So the system is unobservable whenever 2 21 1 2 24 2 0c c c c is satisfied.
b.
If 2 1c the condition becomes
2
1 14 2 0c c which as two solutions 1 0.5858c and
1 3.4142c .
27.
The A and C matrices for the system represented in cascade form is
Az
-20 1 0
0 -13 1
0 0 -5
Cz
1 0 0
The characteristic polynomial found from the transfer function of the plant is
s3 + 38s2 + 425s + 1300
From this characteristic polynomial, we can write observer canonical form of the state equations. The
A and C matrices of the observer canonical form are given below as
12-36 Chapter 12: Design via State Space
Ax
-38 1 0
-425 0 1
-1300 0 0
Cx
1 0 0
To test observability, we write the observability matrices for both systems and show that both
observability matrices have non zero determinants. Using Eq. (12.79),
OMz Observability Matrix of z OMx Observability Matrix of x
1 0 0 1 0 0
-20 1 0 -38 1 0
400 -33 1 1019 -38 1
Det(OMz) 1 Det(OMx) 1
Using Eq. (12.89), we obtain the transformation matrix, P, and its inverse as
P Transformation Matrix z=Px PINV
1 0 0 1.00 0.00 0.00
-18 1 0 18.00 1.00 0.00
25 -5 1 65.00 5.00 1.00
Using the characteristic polynomial given in the problemstatement, the plant’s characteristic equation,
and Eq. (12.67), the observer for the observer canonical system is
Lx Observer for x
562
39575
1498700
Using Eq. (12.92), the observer for the cascade system is found to be
Lz Observer for z
562
29459
1314875
28.
Program:
A=[-20 1 0;0 -13 1;0 0 -5]
B=[0;0;1]
C=[1 0 0]
D=0
poles=roots([1 600 40000 1500000])
L=acker(A',C',poles);
'L'
L'
Computer response:
Solutions to Design Problems 12-37
A =
-20 1 0
0 -13 1
0 0 -5
B =
0
0
1
C =
1 0 0
D =
0
poles =
1.0e+002 *
-5.2985
-0.3508 + 0.4001i
-0.3508 - 0.4001i
ans =
L
ans =
1.0e+006 *
0.0006
0.0295
1.3149
29.
Expanding the plant by partial fractions, we obtain
1 0.008333 0.017857 0.0095238
( )
( 5)( 13)( 20) ( 5) ( 13) ( 20)
G s
s s s s s s
The A and C matrices for the system represented in parallel form is
Az
-5 0 0
0 -13 0
0 0 -20
Cz
1 1 1
The characteristic polynomial found from the transfer function of the plant is
12-38 Chapter 12: Design via State Space
s3 + 38s2 + 425s + 1300
From this characteristic polynomial, we can write the observer canonical form of the state equations.
The A and C matrices of the observer canonical form are given below as
Ax
-38 1 0
-425 0 1
-1300 0 0
Cx
1 0 0
To test observability, we write the observability matrices for both systems and show that both
observability matrices have non zero determinants. Using Eq. (12.79),
OMz Observability Matrix of z OMx Observability Matrix of x
1 1 1 1 0 0
-5 -13 -20 -38 1 0
25 169 400 1019 -38 1
Det(OMz) -840 Det(OMx) 1
Using Eq. (12.89), we obtain the transformation matrix, P, and its inverse as
P Transformation Matrix z=Px PINV
0.2083333 -0.04166667 0.008333333 1.00 1.00 1.00
-3.017857 0.232142857 -0.01785714 33.00 25.00 18.00
3.8095238 -0.19047619 0.00952381 260.00 100.00 65.00
Using the characteristic polynomial given in the problem statement, the plant’s characteristic equation,
and Eq. (12.67), the observer for the observer canonical system is
Lx Observer for x
562
39575
1498700
Using Eq. (12.92), the observer for the parallel system is found to be
Lz Observer for z
10957.29167
-19271.4821
8876.190476
30.
Use Equations (4.39) and (4.42) to find ζ= 0.5912 and n = 135.328. Factoring the denominator of
Eq. (4.22), the required poles are − 80 ± j109.15. We place the third pole 10 times further at − 800.
Multiplying the three closed-loop pole terms yields the desired characteristic equation:
Solutions to Design Problems 12-39
3 2960 146313.746 14650996.915 0s s s
Since
3 2
45
( )
18 95 150
G s
s s s
, the plant in observer canonical form is:
18 1 0 0
95 0 1 ; 0 ; 1 0 0 ;
150 0 0 45
oc oc ocA B C
Using Eq. (12.79),
1 0 0
18 1 0
229 18 1
MocO , which is observable since the determinant is 1.
Since
3 2
45
( )
18 95 150
s
s s s
, the phase-variable form is:
0 1 0 0
0 0 1 ; 0 ; 45 0 0
150 95 18 1
pv pvA B C ;
Using Eq. (12.79),
45 0 0
0 45 0
0 0 45
MpvO , which is observable since the determinant is 91125.
The first column of Aoc contains the negative values of the coefficients of the characteristic equation.
Comparing the first column of Aoc to the desired characteristic equation and using Eq. (12.67), l1 =
960 - 18 = 942; l2 = 146313.746 - 99 = 146214.746; and l3 = 14650996.915 - 162 =
14650834.915.
Hence, Loc = [942 146214.746 14650834.915]. The transformation matrix is,
1
0.0222 0 0
0.4000 0.0222 0
5.0889 0.4000 0.0222
Mpv MocP O O
12-40 Chapter 12: Design via State Space
Transforming back to the original system, Lpv = PLoc = [20.91 2869.17 271556.38]
T
.
31.
a.
Since both systems are in phase variable form it readily follows that:
2det( ) 3 2s s s1I A
and
3 22det( ) 6 11 6s s s sI A
1 2 2
3 1
02 2
( ) ( 2 0
13 2 3 2
s
s
G s s
s s s s
1
1 1 1C I A ) B
2
2
2 2 2 2 3 2
3 2 2
6 11 6 1
6 ( 6)
0
6 (11 6)
( ) ( 6 2 0 0
6 11 6
1
2( 3) 2
6 11 6 3 2
s s s
s s s
s s s
G s s
s s s
s
s s s s s
1
C I A ) B
b.
2 0
0 2
1
M1
1 1
C
O
C A
; ( ) 2rank M1O System 1 is observable
2
6 2 0
0 6 2
12 22 6
2
M 2 2
2
2 2
C
O C A
C A
; 2( ) 2rank MO System 2 is not observable
32.
The open-loop transfer function of the plant is T(s) = C (sI-A)
-1
B =
2
2
2
s
s s
.
Using Eqs. (12.115), the closed-loop state equations with integral control are:
Solutions to Design Problems 12-41
1
1 1
2 1 2 2 2
1 1 0 0
2 0 ; 1 1 0
1 1 0 1
e
N N
N
x
x x
x k k k x r y x
x x
x
The characteristic polynomial is: 3 22 2 1( 1) ( 2) 2e es k s k k k s k
The desired characteristic polynomial is calculated from the desired transient
response stated in the problem. Also, the third pole will be placed to cancel the
zero at -2. Hence, the desired characteristic polynomial is:
2 3 2( 2)( 13.335 166.325) 15.335 192.995 332.65s s s s s s
12-42 Chapter 12: Design via State Space
Equating coefficients of the characteristic polynomials yields,
2 1166.325, 16.335, 12.335ek k k
33.
The open-loop transfer function of the plant is T(s) = C(sI-A)-1B =
2
s 3
s 7s 10
.
Using Eqs. (12.115), the closed-loop state equations with integral control is
1
1 1
2 1 2 2 2
2 1 0 0
-(5 ) 0 ; 1, 1, 0
1 1 0 1
e
N N
N
x
x x
x k k k x r y x
x x
x
The characteristic polynomial is
s3 + (k2 + 7)s
2 + (2k2 + k1 + ke + 10)s + 3ke
The desired characteristic polynomial is calculated from the desired transient response stated in the
problem. Also, the third pole will be placed to cancel the zero at -3. Hence, the desired characteristic
polynomial is
2 3 2( 3)( 16 183.14) 19 231.14 549.41s s s s s s
Equating coefficients of the characteristic polynomials yields,
ke = 183.137, k2 = 12, k1 = 14.003
SOLUTIONS TO DESIGN PROBLEMS
34.
a. The open loop characteristic polynomial is
3 2 3 2
2 1 0( 0.004)( 0.19)( 0.66) 0.854 0.1288 0.0005016s s s s s s s a s a s a The desired
characteristic polynomial is
2 3 2 3 2
2 1 0
1
( ) ( 0.5) 0.854 0.1288 0.0005016
15
s s s s s s d s d s d
So we have
1 0 0 0.0017204k d a
2 1 1 0.05769k d a
3 2 2 0.2207k d a
Solutions to Design Problems 12-43
b.
>> A = [0 1 0; 0 0 1; -0.0005016 -0.1288 -0.854];
>> B=[0;0;1];
>> C = [0.000078 0.00414 0.01];
>> K=[0.0017204 -0.05769 -0.2207];
>> eig(A-B*K)
ans =
-0.49995739538301
-0.06584435043216
-0.06749825418482
35.
a. By inspection the phase variable form of this system results in
0 1 0
0 0 1
0.000078 0.01265 0.5250
A ;
0
0
1
B ;
4 4[ 0.01197 10 0.958 10 0]C ; 0D
b. The characteristic equation ofthe closed loop system will have 3 poles. Two of them will be
placed at 1
4 1
sec
300sec 75
to satisfy the settling time spec. The third pole will be placed farther
to the left of the latter, say at -0.05. The desired characteristic polynomial is
2 3 2( 0.0133) ( 0.05) 0.0766 0.00150689 0.0000088445 0s s s s s
The closed loop matrix is
12-44 Chapter 12: Design via State Space
1 2 3
0 1 0
0 0 1
(0.000078 ) (0.01265 ) (0.525 )k k k
A BK
Which will result in a characteristic polynomial
3 23 2 3(0.525 ) (0.01265 ) (0.000078 ) 0s k s k s k
Equating coefficients for both polynomials we get 3 0.4484k , 2 0.01114311k and
51 6.91555 10k
c.
>> A=[0 1 0; 0 0 1; -0.000078 -0.01265 -0.525];
>> B=[0;0;1];
>> C=[-0.01197e-4 -0.9580e-4 0];
>> D=0;
>> K=[-6.91555e-5 -0.01114311 -0.4484];
>> eig(A-B*K) %check
ans =
-0.0133
-0.0133
-0.0500
>> step(A-B*K,B,C,D)
Solutions to Design Problems 12-45
36.
a. The system’s transfer function
2
83335
( )
10 41665
G s s
s s
1
C( I A) B
b. From the transfer function by inspection
0 1
41665 10
pA ;
0
1
pB ; [83335 0]pC
c. A 20% overshoot gives 0.7 , and
4
11.43n
sT
so the desired characteristic equation
is 2 2 22 16 130.65n ns s s s
Equating gains we have
141665 130.65k and 210 16k giving 1 2[ ] [ 41534.35 6]xK k k
d. The Controllability matrix for the original system is
167 0
0 83335
MzC B AB
The Controllability matrix for the phase variable system is
12-46 Chapter 12: Design via State Space
0 1
1 10
x p P PMC B A B
The transformation matrix is
1
1667 167
83335 0
Mz MxP C C
The set of gains for the original system is:
1 0.0360 0.4991Z xK K P
e. 2det( ( )) 16 130.65zsI A BK s s
f.
>> A=[0 -83.33; 500 -10];
>> B = [166.67;0];
>> C=[0 1];
>> K=[0.0360 -0.499];
>> step(A-B*K,B,C,0)
Solutions to Design Problems 12-47
37.
a. We start by expressing the system in Observable Canonical form. From problem IV, the transfer
function is
2
2
2
83335
( ) 83335
( )
10 41665( ) 10 41665
1
C s sG s
R s s s
s s
, so we can write
1 1
( ) 10 ( ) 83335 ( ) 41665 ( )C s C s R s C s
s s
from which we get the following
diagram:
12-48 Chapter 12: Design via State Space
By direct inspection we can write
0 0
1 1
10 1 0
41665 0 83335
s
x x
E
x x
and
0
1
[1 0]
x
y
x
The characteristic equation for this system is 2 10 41665 0s s with roots at
-5j204.06. To make the observer 10 times faster its poles will be placed at -50j2040.6
The desired characteristic equation is 2 6100 4.167 10 0s s
The error equation is given by
1
2
(10 ) 1
( )
(41665 ) 0
X x x
l
l
e A LC e e which has a characteristic equation
2 1 2(10 ) (41665 ) 0s l s l
By equation coefficients with the desired characteristic equation we obtain
1 100 10 90l and
6
2 4.167 10 41665 4125335l
Solutions to Design Problems 12-49
b.
12-50 Chapter 12: Design via State Space
38.
a. From the solution of Problem III the system can be written in phase variable form as:
0 1 0
0 0 1
0.000078 0.01265 0.5250
zA ;
0
0
1
zB ;
4 4[ 0.01197 10 0.958 10 0]zC ; 0zD
The corresponding observability matrix is:
4 4
4 4
4 4 4
0.012 10 0.958 10 0
0 0.019 10 0.958 10
0.0001 10 0.0121 10 0.491 10
z
Mz z z
2
z z
C
O C A
C A
It is readily verified that the rank of MzO is 3 and the system is observable.
Next we write the system in observable canonical form, for this let
4 4
4 4 2 3
3 2
2 3
0.958 10 0.01197 10
( ) 0.958 10 0.01197 10
( )
0.525 0.01265 0.000078( ) 0.525 0.01265 0.000078
1
C s s s sG s
R s s s s
s s s
From which we can write:
Solutions to Design Problems 12-51
4 4
1 1 1
( ) 0.525 ( ) 0.958 10 ( ) 0.01265 ( ) 0.01197 10 ( ) 0.000078 ( )C s C s R s C s R s C s
s s s
Giving the following state flow diagram.
The system in observer canonical form is
0.5250 1 0
0.01265 0 1
0.000078 0 0
xA ;
4
4
0
0.958 10
0.01197 10
xB ;
[1 0 0]xC ; 0xD
And
1
2
3
(0.5250 ) 1 0
(0.01265 ) 0 1
(0.000078 ) 0 0
x
l
l
l
x xA L C
With a characteristic polynomial
3 21 2 3(0.525 ) (0.01265 ) (0.000078 ) 0s l s l l
12-52 Chapter 12: Design via State Space
The plant has poles at -0.5, -0.0128 and -0.0122. To obtain the tenfold speed increase in the observer
arbitrarily we will place the observer poles 10 times to the left of the plant poles, namely -5, -0.128
and -0.122. The resulting desired polynomial is
3 20.75 0.1406 0.007808 0s s s Comparing the observer and the desired polynomials
we obtain:
1 0.75 0.525 0.225l
2 0.1406 0.01265 0.128l
3 0.007808 0.000078 0.0077l
Or
0.225
0.128
0.0077
xL
The observability matrix for the observable canonical system is:
1 0 0
0.525 1 0
0.263 0.525 1
x
x x x
x x
M
2
C
O C A
C A
The transformation matrix is
4 4 4
4 4 4
4 4 4
14 10 688 10 33480 10
1 10 14 10 688 10
1 10 1 10 14 10
1
Mz mxP O O
Finally
6
2
1.7286 10
379 10
700
z xL OL
Solutions to Design Problems 12-53
b.
Error!
12-54 Chapter 12: Design via State Space
Solutions to Design Problems 12-55
12-56 Chapter 12: Design via State Space
39.
a. As per equation (12.115) in the text the augmented system is:
1 1
2 2
1
1 2
2
1 2
0
0
0
1
0 83.33 166.67 166.67 0
500 10 0 0 0
10 1 0
166.67 (83.33 166.67 ) 166.67
e
N N
e
N
e
x x
k
x x r
x x
x
k k k
x r
x
k k k
A BK B
C
1
2
0
500 10 0 0
0 1 0 1N
x
x r
x
1
20
N
x
y x
x
C
The characteristic equation for the system above is:
Solutions to Design Problems 12-57
1 2
3 2
1 1 2
166.67 83.33 166.67 166.67
det( ) 500 10 0
0 1
(10 166.67 ) (1666.7 41665 83335 ) 83335
e
e
s k k k
s s
s
s k s k k s k
augI A
The desired polynomial for 0.5secsT and %OS=20% with an extra far away pole is:
2 3 2( 16 130.65)( 50) 66 930.6 6532.5s s s s s s
Equating coefficients in both polynomials we have:
1 110 166.67 66 0.336k k
1 2 21666.7 41665 83335 930.6 0.4955k k k
83335 6532.5 0.0783884e ek k
b.
12-58 Chapter 12: Design via State Space
40.
a. We start by verifying that the system is controllable. For a system of this order it is best
to use MATLAB or other computing tools:
>>A=[-5 0 0 0 0; 0 0 1 0 0; -10.5229 -1066.67 -3.38028 23.5107 0; 0 993.804 3.125 -23.5107 0; 0 0
0 10 -10];
>>B = [5;0;0;0;0];
>>C = [0 0 0 1.2331e5 0];>>D=0;
>> rank(ctrb(A,B))
ans =
5
So the system is controllable.
We also use MATLAB to find the corresponding transfer function:
>> [n,d]=ss2tf(A,B,C,D)
n =
1.0e+010 *
Solutions to Design Problems 12-59
0 0.000000000000000 0.000000000000000 -0.002027466873438 -
0.665044169115073 -6.447695003806981
d =
1.0e+004 *
0.000100000000000 0.004189098000000 0.152603651149600 1.914775683863999
7.933055056779999 8.565653331000014
We use this information to express the system in state variable form:
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
85650 79330 19150 1526 41.89 1
x u
xx
64476950038.1 6650441691.2 20274668.73 0 0 xy x
Now we find the desired characteristic polynomial using a dominant pole approximation.
Assuming a pair of complex conjugate dominant poles
4
2s
n
T
s, or 2n . The 10%
OS requirement corresponds to a 0.6 damping factor. Therefore the dominant poles can
be included in a 2 2 22 4 11.11n ns s s s . The must be a total of 5 poles so
arbitrarily we will add three poles 10 times to the left of the dominant poles, namely
3( 20)s . Thus the desired polynomial is given by:
2 3
5 4 3 2
( ) ( 4 11.11)( 20)
64 1451.11 13466.77 45335.47 88903.11
D s s s s
s s s s s
Under state feedback, the phase variable system will become
1 2 3 4 5
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
(85650 ) (79330 ) (19150 ) (1526 ) (41.89 ) 1x x x x x
x u
k k k k k
xx
With a characteristic equation given by:
5 4 3 2
5 4 3
2 1
det( ( ))
(41.89 ) (1526 ) (19150 )
(79330 ) (85650 )
x x x
x x x
x x
sI A B K
s k s k s k s
k s k
12-60 Chapter 12: Design via State Space
We now equate the coefficients of this equation with those of the desired polynomial to get:
1 2 3 4 5 3253.11 33994.53 5683.28 74.89 22.11x x x x x xK k k k k k
This vector is now transformed to the original coordinate system using MATLAB:
>> Ax=[0 1 0 0 0; 0 0 1 0 0; 0 0 0 1 0; 0 0 0 0 1;-8.565653331000014e4 -
7.933055056779999E4 -1.914775683863999e4 -0.152603651149600E4 -
0.004189098000000E4];
>> Bx = [0;0;0;0;1];
>> P=ctrb(A,B)*inv(ctrb(Ax,Bx));
>>Kz=Kx*inv(P);
4.422 110.04 16.93 1.75 0.14K
b. The simulation is performed with the following commands:
>>Af=(A-B*Kz)
>>step(Af,B,C,D)
Solutions to Design Problems 12-61
41.
a. The controllability matrix for the original system is
0 61.7325 2170.9283 34794.0908
0 61.7325 2170.9283 13187.7158
61.7325 2170.9283 34794.0908 237593.3651
61.7325 2170.9283 13187.7158 997418.2728
2 3
MOC B AB A B A B
121.779 10 0MOC so the system is controllable
b. The original systems characteristic equation is
4 3 2det( - ) s 35.1667 1023.07 12308 0s s s sI A
It follows by inspection, that the phase variable representation of the system is:
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 12308.345 1023.07 35.1667 1
uPx x
c. The controllability matrix of the phase variable representation is
2 3
0 0 0 1
0 0 1 35.1667
0 1 35.1667 213.6268
1 35.1667 213.6268 16157.1015
MP P P P P P P PC B A B A B A B
The transformation matrix is
21606.375 0 61.7325 0
0 0 61.7325 0
0 21606.375 0 61.7325
0 0 0 61.7325
-1
MO MPP = C C
d. Using state feedback the closed loop phase state representation has an CLA matrix
12-62 Chapter 12: Design via State Space
1 2 3 4
0 1 0 0
0 0 1 0
-
0 0 0 1
( 12308.345) (k 1023.07) ( 35.1667)P P P Pk k k
CL P P PA = (A B K ) =
which results in a characteristic equation
4 3 24 3 2 1det( ) ( 35.1667) ( 1023.07) ( 12308.345)P P P Ps s k s k s k s kCLI A
Equating coefficients with the desired pole positions
4 4 3 2( ) ( 10) 40 600 4000 10000D s s s s s s
We get 1 2 3 4 10000 8308.34 423.07 4.8333P P P Pk k k kPK
e.
0.4628 7.3161 0.3845 0.4628-1O PK = K P
42.
a. The observability matrix is obtained from
Α
Α
1 0 0
0 1 0
52.6532 4.9353 2768.1557
MZ
2
C
O C
C
2768.1557 0MZO
so the system is observable.
b. The original system’s characteristic equation is
3 2det( ) s 4.9964 52.9546 0.1428sI s sA . The observer canonical form of
the system can be written by inspection, resulting in
4.9964 1 0
52.9546 0 1
0.1428 0 0
XA
Solutions to Design Problems 12-63
1 0 0xC
c. The observability matrix for the observer canonical form representation is
1 0 0
4.9964 1 0
27.9906 4.9964 1
x
MX x x
2
x x
C
O C A
C A
6 5 4
1 0 0
4.9964 1 0
1.4017 10 2.2073 10 3.6125 10
-1
MZ MXP = O O
d. We write
1
2
3
( 4.9964) 1 0
( 52.9546) 0 1
( 0.1428) 0 0
x
x
x
l
l
l
x x xA - L C
From which it follows that
3 21 2 3det( ( - )) s (4.9964 ) (52.9546 ) ( 0.1428)x x xs l s l s lx x xI A L C
Comparing with the specified polynomial 3 2( ) 30 316 1160D s s s s we get
25.0036
263.0454
1160.1428
xL
e.
25.0036
138.1174
0.4133
z xL = PL =
12-64 Chapter 12: Design via State Space
43.
The modified Simulink model is shown below. As could be seen, two state feedback amplifiers
were added: an angle feedback amplifier with a gain of 12 and a speed feedback amplifier with a
gain of 0.1. The settings of the rate feedback amplifier and the PID block (configured as a PD
controller) were kept the same.
As could be seen from the scope graph, shown below, using state feedback amplifiers with
appropriate gains improved the impulse response of the angle control loop if compared the
response obtained in problem IX-z: The amplitude of the pulse here has been reduced to 0.17
radians (less than half of that in IX-z) and it goes down to zero in 0.05 rather than in 0.2 seconds.
l
Solutions to Design Problems 12-65
44.
Dividing all terms in the numerator and denominator of GP (s) by s
3
, we get:
2 3
2 3
1 1.2 12500
( ) 250* ( )
( ) 250*
8.1 62003 31250( ) ( )
1
P
C s Y s s s sG s
M s M s
s s s
Cross-multiplying, and combining terms of like powers gives:
2 3
1 1 1ˆ ˆ ˆ ˆ( ) ( ) 8.1 ( ) 1.2 ( ) 62003 ( ) 12500 ( ) 31250 ( )Y s M s Y s M s Y s M s Y s
s s s
Thus, the plant may be represented as shown below in Figure 1a. Identifying the state variables as
the outputs of the integrators, we have:
8.1 1 0 1
ˆ ˆ ˆx Ax B 62003 0 1 x 1.2
31250 0 0 12500
m m ;
ˆˆ 1 0 0 xy
12-66 Chapter 12: Design via State Space
Now form the difference between the plant’s actual output, y, and the observer’s estimated
output, ŷ , and add the feedback paths from this difference to the derivative of each state variable.
The result is shown in Figure 1b
Figure 1
Solutions to Design Problems 12-67
Next find the characteristic polynomial. From Eqs. (12.64) and (12.66), the observer
error is:
1
2
3
(8.1 ) 1 0
( ) (62003 ) 0 1 ;
(31250 ) 0 0
l
l
l
x x xe A LC ee
Using equation 12.65, we obtain the characteristic polynomial.
3 21 2 3(8.1 ) (62003 ) (31250 )s l s l s l
The dominant poles of the output, c(t), in the proportionally controlled loop of problem
IX-z were – 34.14 ± j111.44. For the observer to respond 10 times faster, its poles
should be at:
– 341.4 ± j1114.4. Select the third pole to be 10 times the real part of the dominant
observer poles, or – 3414. Hence, the desired characteristic polynomial is:
2 3 2( 3414)( 68.28 12418.9) 3482.3 245526.8 42398034s s s s s s
Equating the two polynomials gives:
1 2 33473.9; 183523.8; 42366784;l l l
Interested students can use MATLAB (or any other software) to simulate the observer
and check its response to a step and/or a ramp input.
45.
a.
In chapter 4 we found that the open loop transfer function of this system is type 0. Thus for zero
steady state error, integral control is required. Also as the open loop transfer function has a zero at -
0.02, and this zero will appear as a zero of the closed loop transfer function if not eliminated, it will
be cancelled with a closed loop pole. The 10% overshoot requirement corresponds to a
0.6 damping factor. The settling time requirement correspond to a second order term with
4
0.04n
sT
or 0.0667n . The desired closed loop polynomial is:
12-68 Chapter 12: Design via State Space
2 4 3 2( 0.08 0.0044)( 0.02)( 0.5) 0.6 0.056 0.003088 0.000044s s s s s s s s
where a far away pole was arbitrarily added at -0.5.
Following equation (12.115a) in the text, the closed loop system matrix A is given by
1 2 3
1 2 3
1
0
0.04167 0 0.0058 5.2 5.2
0.0217 0.24 0.0058 5.2 5.2
0 100 2.4 0 0
0 0 1 0
(0.04167 5.2 ) 5.2 (0.0058 5.2 ) 5.2
0.0217 5.2 0.24 5.2
e
CL
e
e
k
k k k k
k k k k
k
A BK B
A
C
2 30.0058 5.2 5.2
0 100 2.4 0
0 0 1 0
ek k k
The corresponding characteristic polynomial is:
4 3 2
1 2 1 2 3
1 2 3
(2.68167 5.2 5.2 ) (0.1060088 13.728 12.583844 520 )
(0.01241932 2.9952 0.2492256 10.3844 520 ) 10.3844e e
s k k s k k k s
k k k k s k
Equating the coefficients of this polynomial with those of the desired polynomial and solving
simultaneous equations one gets
1 2 3 0.004174 0.39615 0.0096k k kK and
64.237125 10ek
Solutions to Design Problems 12-69
b.
46.
We’ll start by finding the key parameters needed to meet specifications:
12-70 Chapter 12: Design via State Space
The damping ratio, for a 4.32% overshoot is:
2 2 2 2
ln(% /100) ln(4.32 /100)
0.707
ln (% /100) ln (4.32 /100)
OS
OS
The natural frequency, n, for a settling time = 4 sec at the above is:
4 4
4 1.414 rad/sec
0.707
s n
n n
T
The coordinates of the required dominant poles, – n ± jd = – 1 ± j 1.
The open-loop transfer-function of the “plant” is given by equation (3.73):
-1
( )
( ) (sI )
( )
Y s
G s
U s
C A B D
Here:
0 20 40 20 40
(sI )
0 0.2491 0.0191 0.2491 0.0191
s s
s s
A
-1
2 2
20 40 0.0191 0.2491
adj
0.2491 0.0191 40 20adj(sI )
(sI )
det(sI ) 20.0191s 10.346 20.0191s 10.346
s s
s s
s s
A
A
A
Hence:
2 2
0.0191 0.2491 0
0 0.06154
40 20 1000( ) 61.54 (s + 20)
( )
( ) 20.0191s 10.346 20.0191s 10.346
s
sY s
G s
U s s s
Now, we write the closed-loop state equations for the system with integral control in a form similar to
that given in the text by equations 12.115a and 12.115b
(Figure 12.21):
1 2
20 40 0 0
0.2491 1000 0.0191 1000 1000 0
0 0.06154 0 1
a a
e
N N
I I
k k K r
x x
And the output equation is given by:
( ) ( ) 0 0.06154 0
a
N
I
y t v t
x
Solutions to Design Problems 12-71
1 2
0 0 20 40 0
(sI ) 0 0 0.2491 1000 0.0191 1000 1000
0 0 0 0.06154 0
e
s
s K K K
s
A
1 2
20 40 0
0.2491 1000 0.0191 1000 1000
0 0.06154
e
s
K s K K
s
The characteristic polynomial for that system is:
3 22 1 2(20.0191 1000 ) ( 40000 20000 61.54 10.346) 1230.8e es K s K K K s K
The desired characteristic polynomial (with the third pole placed to cancel the zero at -20) is:
3 222 42 40s s s
Equating coefficients of the characteristic polynomials yields:
K1 = 0.000249, K2 = 0.001981, Ke = 0.0325.
Substituting these values into the state equations for the system yields:
20 40 0 0
0.0001 2 32.5 0
0 0.06154 0 1
a a
N N
I I
r
x x
( ) ( ) 0 0.06154 0
a
N
I
y t v t
x
To check our assumptions, we use Eq. (3.73) and MATLAB to find the closed-loop transfer function
of the system to be:
12-72 Chapter 12: Design via State Space
3 2 2 2
( ) 2 40 2( 20) 2
( )
( ) 22 42 40 ( 2 2)( 20) 2 2
Y s s s
T s
R s s s s s s s s s
The following MATLAB file was written to plot the step response.
A=[-20 -40 0; 0.0001 -2.000 32.5; 0 -0.06154 0];
B= [0; 0; 1];
C = [ 0 0.06154 0];
D = 0;
[num, den]= ss2tf(A,B,C,D,1);
T = tf(num, den); %T is the closed-loop TF
step (T);
The characteristics displayed on the step response shown below indicate that the desired transient and
steady-state response requirements are met.
To find, analytically, the steady-state error for a unit step input, we apply Equation (7.96) to the state
equations to obtain:
1
20 40 0 0
( ) 1 0 0.06154 0 0 0001 2 32.5 0 0
0 0.06154 0 1
e .
Thus, the system behaves like a Type 1 system.
Solutions to Design Problems 12-73
47.
The phase-variable form representation of the system is
1 1 6
6
2 2
0 1 0
137.2 10 ( )
196 10 0.0024 1
x x
u t
x x
1
2
1 0
x
y
x
The desired closed-loop poles are obtained from
4
200s
n
T from which we get
0.04n , so
2 2 22 0.04 0.0016n ns s s s the roots of which are
0.02 0.03464j . Since integral control will be used a far-away pole is added to the
polynomial, resulting in
2 3 2( 0.04 0.0016)( 0.1) 0.14 0.0056 0.00016s s s s s .
Following the analysis in Section 12.8, the system with feedback can be represented by
12-74 Chapter 12: Design via State Space
1 1
1 26
2 2
1
6
1 2 2
0 1 0 0 0
196 10 0.0024 1 1 0
11 0 0
0 1 0 0
196 10 0.0024 0
11 0 0
e
N N
e
N
x x
k k K
x x r
x x
x
k k K x r
x
1
21 0 0
N
x
y x
x
The characteristic equation for this system is found by inspection to be
3 2 62 1(0.0024 ) (196 10 ) es k s k s K . Equation coefficients with the desired
polynomial we get 1 0.005404k , 2 0.1376k and 0.00016eK . We substitute
these values into the equation above to get:
1 1
2 2
0 1 0 0
0.0056 0.14 0.000160
1 0 0 1N N
x x
x x r
x x
1
21 0 0
N
x
y x
x
The step response of this system results in:
Solutions to Design Problems 12-75
ONLINEFFIRS 11/25/2014 13:29:37 Page 1
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Library of Congress Cataloging-in-Publication Data
Nise, Norman S.
Control systems engineering / Norman S. Nise, California State Polytechnic University, Pomona. — Seventh edition.
1 online resource.
Includes bibliographical references and index.
Description based on print version record and CIP data provided by publisher; resource not viewed.
ISBN 978-1-118-80082-9 (pdf) — ISBN 978-1-118-17051-9 (cloth : alk. paper)
1. Automatic control–Textbooks. 2. Systems engineering–Textbooks. I. Title.
TJ213
629.8–dc23
2014037468
Printed in the United States of America
10 9 8 7 6 5 4 3 2 1
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